Answer to Question #73653 in Calculus for Shiva

Question

Find the point on the ellipse x^2/4 +y^2 =1, that is nearest to the origin.

Answer 1

We are dealing with the equation of the eclipse.
1) Rewrite x,y in polar coordinates: x = r * Cos[ a ], y = r * Sin[ a ].
We are looking for minimal values of r^2 ( squared distance from the origin ).
2) The equation now take a form: r^2 * ( Cos[a]^2 / 4 + Sin[a]^2 ) = 1,
so r^2 = 1 / ( Cos[a]^2 + 4 * Sin[a]^2 ) = 4 / ( 1 + 3 * Sin[a]^2 ) and is defined and differentiable for any angles from 0 to 2Pi.
3) To find the minimum differentiate r^2. d(r^2) / da = - 24 * Sin[a] * Cos[a] / ( 1 + 3 * Sin[a]^2 )^2. Its derivative has zeroes at 0, Pi/2, Pi and 3Pi/2. The minimal value of r^2 should be at one or several points from this list.
4) r^2( 0 ) = 4, r^2( Pi / 2) = 1, r^2( Pi ) = 4, r^2( 3Pi/2 ) = 1. So we see that the same minimal value take place at Pi/2 and 3Pi/2, which corresponds to (0,1) and ( 0, -1 ).

Answer: ( 0,1 ) OR ( 0, -1 ).